Question

The ostwald process is used commercially to produce nitric acid, which is, in turn, used in many modern chemical processes. in the first step of the ostwald process, ammonia is reacted with oxygen gas to produce nitric oxide and water. what is the maximum mass of h2o that can be produced by combining 54.0 g of each reactant?

Answer 1

Answer:

36.45 g

Explanation:

The balanced reaction given is:

2NH₃(g) + (5/2)O₂(g) → 2NO(g) + 3H₂O(l)

The molar masses of the interest compounds are:

NH₃ = 17.0 g/mol

O₂ = 32.0 g/mol

H₂O = 18.0 g/mol

First, let's find out which of the reactants will be totally consumed, the limiting, and which one is in excess.

By the reaction, the stoichiometry is:

2 moles of NH₃ -------------- 5/2 moles of O₂

Converting by mass (multiplying the molar mass by the number of moles), and supposing that all the ammonia will be consumed:

34 g of NH₃ ------------- 80 g of O₂

54.0 g          ------------- x

By a simple direct three rule:

34x = 4320

x = 127 g of O₂

So, it'll be needed more oxygen than we have, so, oxygen must the limiting reactant.

The stoichiometry between oxygen and water is:

5/2 moles of O₂ -------------------- 3 moles of H₂O

Converting by mass:

80 g of O₂ ------------------------ 54 g of H₂O

54 g           ------------------------ y

By a simple direct three rule:

80y = 2916

y = 36.45 g of H₂O

Answer 2

The maximum mass of water produced when 54.0 g of each reactant are combined in the first step of the Ostwald process is 45.6 g.

Let's consider the balanced equation for the first step of the Ostwald process.

2 NH₃ + 2 O₂ ⇒ NO + 3 H₂O

Considering the balanced equation, the theoretical mass ratio (TMR) of NH₃ to O₂ is:

$TMR = \frac{mNH_3}{mO_2} = \frac{34.06gNH_3}{64.00 gO_2} = \frac{0.5322gNH_3}{1gO_2}$

Considering the given data, the experimental mass ratio (EMR) of NH₃ to O₂ is:

$EMR = \frac{mNH_3}{mO_2} = \frac{54.0gNH_3}{54.0 gO_2} = \frac{1.00gNH_3}{1gO_2}$

We need 0.5322 g of ammonia per gram of oxygen and we have 1.00 g of ammonia per gram of oxygen. Thus, ammonia is the excess reactant and oxygen is the limiting reactant.

The theoretical mass ratio of O₂ to H₂O is 64.00:54.06. The mass of water produced from 54.0 g of oxygen is:

$54.0 g O_2 \times \frac{54.06gH_2O}{64.00gO_2} = 45.6 g H_2O$

The maximum mass of water produced when 54.0 g of each reactant are combined in the first step of the Ostwald process is 45.6 g.

You can learn more about the limiting reactant here: https://brainly.com/question/14225536